lecture_13 - Maximum flow To remember max flow = min cut...

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Unformatted text preview: Maximum flow To remember: max flow = min cut Reading s Dasgupta: Chapter 7.2 s Cormen et al. : Chapter 26.1-2 ssignment 1 Assignment 1 s Average 29.4 / 50 (expectation: 30 – 35) s Max 45 / 50 s Difficulty rating: ~4 (expectation: 4) Mid-term (Friday 10:40 – 11:35) s Divide and conquer, graph algorithms, greedy algorithms, dynamic programming Networks and Flow s A flow network or simply network G = (V, E, c) is a directed graph in which every edge (u,v) has a capacity c(u,v) ≥ 0. G has two distinguished vertices: a source s and a sink t . s Example : v v 12 t v 4 v 2 3 1 s 16 9 14 4 20 7 10 13 Imagine an edge as a pipe. Capacity = liter per min. Flow A flow f on G assigns a real value to every edge in G (i.e., f: E b R) that satisfies two constraints: s Capacity constraint: For every edge (u,v) ∈ E, f (u,v) ≤ c(u,v). s Conservation constraint : zero or positive For every vertex v ∈ V – {s, t}, . y) f(v, v) f(x, E v) (x, E y) (v, ∑ ∑ ∈ ∈ = total inflow total outflow = v Example b Capacity constraint: For every edge (u,v) ∈ E, f(u,v) ≤ c(u,v). b Flow conservation constraint : For every vertex v ∈ V – {s, t}, . y) f(v, v) f(x, E v) (x, E y) (v, ∑ ∑ ∈ ∈ = t v 4 v 2 v 3 v 1 s 11/ 16 4/ 9 11/ 14 4/ 4 15/ 20 12/ 12 7/ 7 1/ 10 8/ 13 Value (size) of a flow s A flow f on G is a function that assigns a real value to every edge in G, i.e., f: E b R. s Value of the flow f : value(f), size(f), v(f) value(f) = ∑ ( s ,y) ∈ E f( s ,y) = ∑ (v, t ) ∈ E f(v, t ). B Example : value(f) = 19 t v 4 v 2 v 3 v 1 s 11/ 16 4/ 9 11/ 14 4/ 4 15/ 20 12/ 12 7/ 7 1/ 10 8/ 13 19 Maximum flow s Given G=(V, E, c), find a flow of G with the maximum value. s How to solve this problem? s Greedy algorithm : tart with some trivial flow of G, say, an empty flow. b Start with some trivial flow of G, say, an empty flow. b Repeatedly augment the current flow until the flow is maximum. s Questions : b How to systematically augment a flow of G. b How to determine whether the current flow is maximum. A path from s to t. s t v u 5/7 15/27 12/57 red vertical line shows total inflow; green vertical line shows total outflow. A small increase s v u 5/7 15/27 12/57 To increase the value of the flow, we increase the flow out of s by some very small value ε . t + ε Influx problem: total inflow > total outflow Push the excess flow along the path s t v u 5/7 15/27 12/57 + ε This edge hasn’t used up its capacity. Residual capacity = 27 – 15 = 12. Push the excess units out of u through this edge. Propagate the increase s v u 5/7 15/27 12/57 + ε t + ε Terminate the push at the sink s v u 5/7 15/27 12/57 t + ε + ε + ε We don't need to conserve the flow at the sink. How much to push? As much as possible 5/27 Note that ε cannot be larger than the smallest residual capacity on the path....
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This note was uploaded on 03/01/2010 for the course CS 1234 taught by Professor Chan during the Spring '10 term at University of the Bío-Bío.

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lecture_13 - Maximum flow To remember max flow = min cut...

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